Upgrade to access all content for this subject

Using Newton's Method, we wish to find a root of:

$$f(x)=\sin(x)-\tan(x)+1$$

...starting with $x_0=\dfrac{\pi}{4}$. Find $x_1$.

$\dfrac{\pi}{4}-\dfrac{1}{2}$

$\dfrac{\pi}{4}+\dfrac{\sqrt{2}-4}{\sqrt{2}}$

$\dfrac{\pi}{2}-\dfrac{\sqrt{2}}{\sqrt{2}-2}$

$\dfrac{\pi}{4}-\dfrac{\sqrt{2}}{\sqrt{2}-4}$